Manifold #14. Quotient Spaces매니폴드 #14. 몫공간

These are my notes on XylyXylyX’s What is a Manifold? lectures. This post covers lesson 14 of the playlist, Quotient Spaces. In the previous post we looked at fiber bundles and the tangent bundle. This time we study the quotient space, a way to build new topological spaces.

Why quotient spaces?

Many of the manifolds we will build are defined as quotient spaces: the cylinder, the Möbius strip, the projective plane, and the sphere. If you have seen the pictures in general relativity where pieces of spacetime are glued together along lines, the exact definition of that gluing is a quotient space. We need quotient spaces to understand gluing properly, and along the way we reuse the topological ideas from earlier lectures.

Equivalence relations

We start with the real line $\R$ with its usual topology. For two points $x, y$ we write $x \sim y$ when there is an integer $k$ with $y = x + 2\pi k$. Pick a point $x$ and the points $x \pm 2\pi$, $x \pm 4\pi$, … continue forever in both directions.

For such a relation to be an equivalence relation it needs three properties.

1. Reflexivity: $x \sim x$. Take $k = 0$.
2. Symmetry: if $x \sim y$ then $y \sim x$. Replace $k$ by $-k$.
3. Transitivity: if $x \sim y$ and $y \sim z$ then $x \sim z$. The sum of two integers is an integer.

Being equivalent is different from being equal. $x$ and $x + 2\pi$ are different points, and we only agreed to treat them as the same thing.

Equivalence classes and partitions

The set of all points equivalent to $x$ is called the equivalence class, written $[x] = \{\, x + 2\pi k \mid k \in \Z \,\}$. It is a countably infinite set, and it holds only points spaced at regular intervals, not the whole line.

When the relation is an equivalence relation, every point lies in exactly one equivalence class. So the classes cut the line into disjoint pieces, which we call a partition. In this example every point of the interval $[0, 2\pi)$ has one class, and $2\pi$ is left out because it lies in the same class as $0$. Notation for the relation varies between books, and you see forms such as $x \sim y$ and $x \sim_R y$.

The quotient space and the canonical projection

The set of all equivalence classes is called the quotient space and written $\R/\!\sim$. Its elements are classes, not points, so it is a set of sets. The map that sends each point to its own class is $p : \R \to \R/\!\sim$, $p(x) = [x]$, called the canonical projection. Once the equivalence relation is given, this map is determined uniquely, which is why it is called canonical (natural).

So far this is only a set with no topology. The interval $[0, 2\pi)$ has the right number of points, but we cannot say that this interval is the quotient space or that it is homeomorphic to it. The topology is still to be chosen. First, move a point. Drag the point $x$ and every point of its class lights up, and in the quotient space below they gather into one point.

Q. Are the points of one equivalence class really the same point?
A. On the line they are different points. It is only in the quotient space that they become a single point (an element). That is why we distinguish “equivalent” from “equal”.

The topology of the quotient space: the final topology

There is one principle for making a new topology: start from a topology and a map you already know, and make the map continuous. For fiber bundles we had a projection $\pi : E \to B$, knew the topology of $B$, and did not know that of $E$. So we gave $E$ the coarsest topology that makes $\pi$ continuous, and called it the initial topology.

This time it is the other way around. For $p : \R \to \R/\!\sim$ we know the topology of the domain $\R$ and the target has none. So we give the target the finest topology that makes $p$ continuous. It is called the final topology, or the quotient topology.

Preimages preserve unions and intersections, so this collection obeys the four rules of a topology.

Why “finest”? For continuity the preimage of every open set of the target must be open in the domain. The more open sets the target has, the harder this condition is. So the topology that accepts as many open sets as the condition allows is the collection above. The coarsest, trivial topology also makes $p$ continuous, but it tells us nothing. A topology that is too fine, such as the discrete topology, fails. If a single point $[x]$ were open, its preimage $x + 2\pi \Z$ would have to be open in $\R$, and it is not.

Q. The initial topology is the coarsest and the final topology is the finest. Why do they go in opposite directions?
A. Both pick the most useful topology among those that make the map continuous. For the initial topology, adding open sets to the domain makes the condition easier, so we add as few as possible. For the final topology, adding open sets to the target makes the condition harder, so we add as many as we can.

Quotient maps and saturated sets

A map $f : X \to Y$ is an open map if the image of every open set is open, and a closed map if the image of every closed set is closed. Continuity is a different notion. It is a condition on preimages, so a map can be continuous without being open, and the other way round.

A quotient map is a map $p : X \to Y$ with three properties.

1. It is surjective.
2. It is continuous.
3. If the preimage $p^{-1}(U)$ of $U \subset Y$ is open, then $U$ is open.

The third condition is the reverse direction of continuity, and together they say that $U$ is open exactly when $p^{-1}(U)$ is open. The canonical projection with the quotient topology satisfies all three.

A saturated set is a notion used in this setting. A subset $A$ of $X$ is saturated when it is exactly a union of equivalence classes, that is, when it contains each class whole without cutting any. Its use is this: for a quotient map $p$, the image of an open saturated set is open. A saturated set $A$ satisfies $p^{-1}(p(A)) = A$, and since $A$ is open, $p(A)$ is open by the definition of the quotient topology. (This reasoning is added for understanding and was not covered in the lecture.) For an open set that is not saturated, the image may fail to be open. So a quotient map need not be an open map.

Q. Is there an example of a quotient map that is not open?
A. Collapsing the whole interval $[0, 1]$ of the line to one point is one. The image of the open set $(-1, \tfrac12)$ is not open, because its preimage is $(-1, 1]$, which is not open in $\R$. (This example is added for understanding and was not covered in the lecture.)

A quotient space is not a look-alike of the original

A homeomorphism preserves every topological property. Hausdorff stays Hausdorff, connected stays connected, and the dimension stays the same. A quotient map does not. Connectedness survives because a continuous map preserves it, but properties such as simple connectedness are not guaranteed. The line is simply connected, and as we will see, its quotient, the circle, is not. Separation properties and the dimension can also differ from the original. Keep in mind that a quotient space can be a very different space from the original.

How a line becomes a circle

To simplify the computation we change the spacing from $2\pi$ to $1$. On the line $\R$ declare $x \sim y$ when $y - x \in \Z$, which gives the quotient space $\R/\!\sim$. We show that it is homeomorphic to the circle $S^1$. Give the circle the subspace topology from the plane, and define $Q : \R \to S^1$ by $Q(t) = (\cos 2\pi t, \sin 2\pi t)$.

1. $Q(s) = Q(t)$ means exactly $s - t \in \Z$. So $Q$ gives a one-to-one correspondence $\tilde{Q} : \R/\!\sim \;\to S^1$ sending one class to one point of the circle, defined by $\tilde{Q}([t]) = Q(t)$.
2. $Q$ is a continuous surjection. An open interval goes to an open arc, and one checks that if the preimage of an open set $O$ of the circle is open in $\R$, then $O$ is open. Saturated sets are used in this check. So $Q$ is also a quotient map.
3. The canonical projection $p$ and $Q$ are two quotient maps that group the points in the same way, so both $\tilde{Q}$ and its inverse are continuous. Hence $\tilde{Q}$ is a homeomorphism.

The proof is long, but the point is clear. When people say “fold the line at integer spacing and you get a circle”, what they really mean is that the quotient space is homeomorphic to the circle. The quotient space itself is a set of sets, while the circle is a subset of the plane, so they are different objects. Many books skip this step.

It also explains why the interval $[0, 1)$ must not be taken as the quotient space. The number of points matches, but the topology differs. On the circle an open set near $0$ also contains points near $1$, while in $[0, 1)$ there are no points below $0$ near $0$.

Changing the equivalence relation

Keep the same line and change only the relation. Declare $x \sim y$ only when both are integers, and let every non-integer be equivalent only to itself. Reflexivity, symmetry, and transitivity all hold, so it is an equivalence relation. Press the “Integers together” button in the toy above.

In this quotient space all the integers gather into one point, and each interval $[n, n+1]$ has both ends attached to that point and becomes a loop. Infinitely many loops meet at one point. This space is nothing like the earlier circle. The line is second countable and so is the circle, but this quotient space is known not even to be first countable at the gathered point. Countability is not a property that quotient maps preserve. Starting from the same line, different relations give completely different spaces.

In the next lecture we use this tool to glue two sides of a square into a cylinder and prove that the quotient space is homeomorphic to the cylinder.